Question: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $12.9$ years; the standard deviation is $2.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living longer than $19.5$ years.
Explanation: $12.9$ $10.7$ $15.1$ $8.5$ $17.3$ $6.3$ $19.5$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $12.9$ years. We know the standard deviation is $2.2$ years, so one standard deviation below the mean is $10.7$ years and one standard deviation above the mean is $15.1$ years. Two standard deviations below the mean is $8.5$ years and two standard deviations above the mean is $17.3$ years. Three standard deviations below the mean is $6.3$ years and three standard deviations above the mean is $19.5$ years. We are interested in the probability of a meerkat living longer than $19.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the meerkats will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $6.3$ years and the other half $({0.15\%})$ will live longer than $19.5$ years. The probability of a particular meerkat living longer than $19.5$ years is ${0.15\%}$.